Solving Pipe and Tank Problems: Finding Emptying Time

This problem involves understanding the concept of work rates for pipes filling and emptying a tank. We are given the filling time for one pipe (Pipe A), the combined filling time when Pipe A and an emptying pipe (Pipe B) work together, and we need to find the emptying time for Pipe B.

The key concept is to represent the work done by each pipe as a rate, which is the fraction of the tank filled or emptied per unit of time (in this case, per hour).

Calculating Individual and Combined Rates

• Rate of Pipe A (filling): Pipe A fills the tank in 12 hours. Its filling rate is the amount of tank filled per hour.
  Rate of A = \(\frac{1}{\text{Time taken by A}} = \frac{1}{12}\) tank per hour.
• Rate of Pipe B (emptying): Pipe B empties the tank in X hours. Its emptying rate is the amount of tank emptied per hour.
  Rate of B = \(\frac{1}{\text{Time taken by B}} = \frac{1}{X}\) tank per hour.
• Combined Rate (filling): When both Pipe A (filling) and Pipe B (emptying) are opened together, the tank is filled in 30 hours. The combined rate is the net amount of tank filled per hour.
  Combined Rate = \(\frac{1}{\text{Time taken when A and B work together}} = \frac{1}{30}\) tank per hour.

Setting Up the Equation

When a filling pipe and an emptying pipe work together, the net rate is the filling rate minus the emptying rate. Since the tank is being filled overall, the filling rate (Pipe A) must be greater than the emptying rate (Pipe B).

Combined Rate = Rate of A - Rate of B

Substituting the rates we found:

\(\frac{1}{30} = \frac{1}{12} - \frac{1}{X}\)

Solving for X (Emptying Time)

Our goal is to find the value of X. We need to rearrange the equation to isolate \(\frac{1}{X}\).

Add \(\frac{1}{X}\) to both sides and subtract \(\frac{1}{30}\) from both sides:

\(\frac{1}{X} = \frac{1}{12} - \frac{1}{30}\)

To subtract the fractions on the right side, we need a common denominator. The least common multiple (LCM) of 12 and 30 is 60.

Convert the fractions to have a denominator of 60:

• \(\frac{1}{12} = \frac{1 \times 5}{12 \times 5} = \frac{5}{60}\)
• \(\frac{1}{30} = \frac{1 \times 2}{30 \times 2} = \frac{2}{60}\)

Now substitute these back into the equation:

\(\frac{1}{X} = \frac{5}{60} - \frac{2}{60}\)

Subtract the numerators:

\(\frac{1}{X} = \frac{5 - 2}{60}\)

\(\frac{1}{X} = \frac{3}{60}\)

Simplify the fraction \(\frac{3}{60}\):

\(\frac{3}{60} = \frac{3 \div 3}{60 \div 3} = \frac{1}{20}\)

So, we have:

\(\frac{1}{X} = \frac{1}{20}\)

Taking the reciprocal of both sides gives us X:

\(X = 20\)

Therefore, Pipe B can empty the tank in 20 hours.

Revision Table: Pipe and Tank Concepts

Additional Information: Understanding LCM in Rate Problems

In pipe and tank or time and work problems, the Least Common Multiple (LCM) of the individual times taken is often used to assume a 'total work' unit. This can sometimes simplify calculations, especially when dealing with multiple pipes or workers.

For this problem, the LCM of 12 (Pipe A's time) and 30 (Combined time) is 60. We can think of the tank having a capacity of 60 units.

• Pipe A's filling rate: \(\frac{60 \text{ units}}{12 \text{ hours}} = 5\) units/hour (filling)
• Combined rate: \(\frac{60 \text{ units}}{30 \text{ hours}} = 2\) units/hour (filling)
• Let Pipe B's emptying rate be 'r' units/hour.

When A and B work together, the net filling rate is A's rate minus B's rate (since B is emptying):

Net Rate = Rate of A - Rate of B

\(2 = 5 - r\)

Solving for r:

\(r = 5 - 2 = 3\) units/hour (emptying)

If Pipe B empties 3 units per hour, the time taken to empty the whole tank (60 units) is:

Time for B = \(\frac{\text{Total Capacity}}{\text{Rate of B}} = \frac{60 \text{ units}}{3 \text{ units/hour}} = 20\) hours.

This method using LCM confirms the result obtained using fractions and provides an alternative way to think about the problem.